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Math Help - Limit of Sequence - max(a,b)

  1. #1
    Junior Member utopiaNow's Avatar
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    Limit of Sequence - max(a,b)

    Prove that if a,b are positive real numbers then \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}.

    My attempt is to say that without loss of generality let a > b, then for sufficiently large n we will have a^n + b^n \approx a^n hence \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a . However this step "for sufficiently large n, we will have a^n + b^n \approx a^n" does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

    Any suggestions would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by utopiaNow View Post
    Prove that if a,b are positive real numbers then \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}.

    My attempt is to say that without loss of generality let a > b, then for sufficiently large n we will have a^n + b^n \approx a^n hence \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a . However this step "for sufficiently large n, we will have a^n + b^n \approx a^n" does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

    Any suggestions would be greatly appreciated.

    As you did, assume wrg that a>b and then "take a out of the parentheses" :

    \sqrt[n]{a^n+b^n}=a\sqrt[n]{1+\left(\frac{b}{a}\right)^n} and remember what happens when we have a geometric sequence with quotient q s.t. |q|<1

    Tonio
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